# Stored energy of a capacitor

Imagine the two terminal of a parallel-plate capacitor are connected to the two terminal of a battery with electric potential difference $$V$$. If the capacitance of the capacitor is $$C$$, and the area of each plate is $$A$$. In this process would the energy lost by the battery and the stored energy of the capacitor be the same or different? Please explain.

Someone pointed that the energy lost by the battery is $$V=\frac{Qd}{\epsilon A}$$ (because the electric potential difference would be used to to raise the potential difference between the plates) and energy gained by the capacitor is $$E=\frac{1}{2}QV$$. And therefore they are different. But I doubt this since the energy should be conserved.

Half of the work done by the battery get stored in the capacitor and half get lost through the resistance as heat. Energy stroed in capacitor $$=\frac12QV=\frac12CV^2$$ Workdone by battery $$=QV=CV^2$$ where $$Q$$ charge flows through the battery and stored on the capacitor. The work lost as heat causes the drift velocity of electrons to be constant through out the circuit.

– user310884
Aug 30, 2021 at 7:29
• In case of charging of a capacitor, Current through the circuit till charging is I=V/R(1-e-Rt).After integrating I²Rdt, power dissipated comes out to be 1/2CV².I am assuming this as a ideal case otherwise some energy can also be lost in radiation also Aug 30, 2021 at 7:42
• How did you get $I= \frac{V}{R(1-e-Rt)}$ ?
– user310884
Aug 30, 2021 at 7:56
• I have written wrong formula .By the way you can get this formula by appyling kirchoff voltage law in a circuit consisting of battery, Resistance and capacitor. Aug 30, 2021 at 8:14

If the capacitor is already charged, its energy is $$E_C=\frac{Q^2}{2C}=\frac{CV^2}{2}.$$ However we cannot say ho much energy the battery has lost while charging, since it depends on the details of the circuit. If the capacitor and the battery are connected by ideal wires, the energy lost by the battery is the same as the energy gained by the capacitor. However, if the circuit contains a resistance, there will be also Joule's heat losses, and more complex circuits may result in other losses (e.g., radiation of EM waves, energy transferred to a mechanical motion of a motor, etc.).

• So, if the circuit is ideal the energy lost by the battery is the same as the energy gained by the capacitor?
– user310884
Aug 30, 2021 at 7:31
• No, the battery loses energy $QV = CV^2$, whereas the capacitor gains energy $\frac{1}{2} C V^2$. The other half is dissipated at the resistance of the wires (irrespective of however small the resistance is). Aug 30, 2021 at 7:58
• @ArchismanPanigrahi Yes, I heard this claim - it is surely correct for a simple RC circuit, but I have not seen a general proof for an arbitrary impedance and/or time dependence. Aug 30, 2021 at 8:40
• The power supplied by the battery is $\int V I(t) dt = V Q_{final} = V (CV) = CV^2$, and the capacitor gains energy $\int_{q=0}^{Q_{final}} V(q) dq = \int_{q=0}^{Q_{final}} \frac{q}{C} dq = \frac{Q_{final}^2}{2C}= \frac{1}{2} CV^2$, and the dissipated energy must be $\frac{1}{2} C V^2$. This holds for any arbitrary time dependence of the current. Alternatively, write down the differential equation for an RC circuit, and solve it to find current as a function of time. You will find that the energy dissipated in the resistor ($\int_0^{\infty} I^2 (t) R dt$) is $\frac{1}{2} C V^2$ Aug 30, 2021 at 9:01
• @ArchismanPanigrahi As I noted in my previous comment, I agree that it is straightforward for a simple RC circuit. Your proof however does not work for an arbitrary impedance - e.g., it may allow for other capacitors in the circuit. Aug 30, 2021 at 9:05